2. Add Two Numbers

You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

Example:

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.

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本题考查链表的基本操作，类似于归并排序的merge过程。 第一版代码如下：

class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2)
    {
        ListNode *ans = new ListNode(0), *tail;
        tail = ans;
        int sum, carry = 0;
        while (l1 && l2) {
            sum = l1->val + l2->val + carry;
            if (sum >= 10) {
                carry = 1;
                sum = sum % 10;
            }
            else
                carry = 0;
            ListNode* ln = new ListNode(0);
            tail->next = ln;
            tail = tail->next;
            tail->val = sum;
            l1 = l1->next;
            l2 = l2->next;
        }
        while (l1) {
            sum = l1->val + carry;
            if (sum >= 10) {
                carry = 1;
                sum = sum % 10;
            }
            else
                carry = 0;
            ListNode* ln = new ListNode(0);
            tail->next = ln;
            tail = tail->next;
            tail->val = sum;
            l1 = l1->next;
        }
        while (l2) {
            sum = l2->val + carry;
            if (sum >= 10) {
                carry = 1;
                sum = sum % 10;
            }
            else
                carry = 0;
            ListNode* ln = new ListNode(0);
            tail->next = ln;
            tail = tail->next;
            tail->val = sum;
            l2 = l2->next;
        }
        if (carry) {
            ListNode* ln = new ListNode(0);
            tail->next = ln;
            tail = tail->next;
            tail->val = 1;
        }
        return ans->next;
    }
};

本代码提交AC，用时40MS。
但是这个版本代码太长了，而且很多重复，不够优雅，翻了翻discuss，有个大神的代码很漂亮，摘录如下：

class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2)
    {
        ListNode *head = NULL, *prev = NULL;
        int sum, carry = 0;
        while (l1 || l2) {
            int v1 = l1 ? l1->val : 0;
            int v2 = l2 ? l2->val : 0;
            sum = v1 + v2 + carry;
            carry = sum >= 10 ? 1 : 0;
            sum %= 10;
            ListNode* ln = new ListNode(sum);
            if (!head)
                head = ln;
            if (prev)
                prev->next = ln;
            prev = ln;
            l1 = l1 ? l1->next : NULL;
            l2 = l2 ? l2->next : NULL;
        }
        if (carry) {
            ListNode* ln = new ListNode(carry);
            prev->next = ln;
        }
        return head;
    }
};

本代码提交AC，用时36MS。
两个版本时间差不多，但是第二个版本看起来就很舒服。